Here \(e\)  runs over the edges of \(Z\)  and \(\deg _Z(e)\)  is the number of faces of \(Z\)  containing \(e\) , see formula (2) in [1]} and formula (8) in [2]}.
In our case, \(L(Z) =-5\)  and \(\chi (Z) = 2\) ; therefore,
\(\mu _1(Z) = \frac{1}{3} + \frac{1}{3e(Z)}, \quad \mu _2(Z) = \frac{1}{2} - \frac{1}{2f(Z)}.\) 
<FIGURE>